# Thread: Help with partial diferentiation (chain rule again)

1. ## Help with partial diferentiation (chain rule again)

The problem I'm working on is for a boundary layer equation for an aerodynamics problem. But I'm having issues working out the partials for it, its been a long while since I've done them and I'm just not getting it to work out. Now for the problem.

I have functions:
$u(x,y)=Uf^{\prime}$ where $f^{\prime}$ is $\frac{\partial f}{\partial \eta}$
$U(x)$
$\eta(x,y)=yg(x)$

I need to find the following partials for my problem:
$\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial^{2}u}{\partial y^{2}}$, $\frac{\partial \eta}{\partial x}$, $\frac{\partial \eta}{\partial y}$, $\frac{\partial^{2}\eta}{\partial y^{2}}$, $\frac{\partial^{2}\eta}{\partial x^{2}}$, $\frac{\partial^{2}\eta}{\partial x\partial y}$

The second order are the ones I know I'm totally missing, the others I may be getting but it'd be nice to get a sanity check. The ones in terms of $\eta$ and x may cancel out when it goes back into my aerodynamics but I'm not positive of that yet.

Thanks for any help, believe me I've given this problem a solid 14 hours of head bashing yesterday before asking for help.

2. Originally Posted by detrepid
The problem I'm working on is for a boundary layer equation for an aerodynamics problem. But I'm having issues working out the partials for it, its been a long while since I've done them and I'm just not getting it to work out. Now for the problem.

I have functions:
$u(x,y)=Uf^{\prime}$ where $f^{\prime}$ is $\frac{\partial f}{\partial \eta}$
$U(x)$
$\eta(x,y)=yg(x)$

I need to find the following partials for my problem:
$\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial^{2}u}{\partial y^{2}}$, $\frac{\partial \eta}{\partial x}$, $\frac{\partial \eta}{\partial y}$, $\frac{\partial^{2}\eta}{\partial y^{2}}$, $\frac{\partial^{2}\eta}{\partial x^{2}}$, $\frac{\partial^{2}\eta}{\partial x\partial y}$

The second order are the ones I know I'm totally missing, the others I may be getting but it'd be nice to get a sanity check. The ones in terms of $\eta$ and x may cancel out when it goes back into my aerodynamics but I'm not positive of that yet.

Thanks for any help, believe me I've given this problem a solid 14 hours of head bashing yesterday before asking for help.
for the first proceeding formally we get

$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left( Uf' \right) =\frac{\partial U }{\partial x}f'+U\frac{\partial f'}{\partial x}$

By the product rule.

Now we need to break each of these down.

$\frac{d U }{d x}f'+U\left(\frac{\partial f^2}{\partial^2 \eta} \right)\frac{\partial \eta}{\partial x}=\frac{d U }{d x}f'+U\left(\frac{\partial f^2}{\partial^2 \eta} \right) \left(y\frac{dg}{dx} \right)$

Try this with some of the others

3. Ok let me post what I've got thus far:
$\frac{\partial u}{\partial x} =\frac{\partial U}{\partial x}f^{\prime}+Uf^{\prime\prime}\frac{\partial \eta}{\partial x}$
$\frac{\partial u}{\partial y}=Uf^{\prime\prime}g$
$\frac{\partial \eta}{\partial y}=g$
$\frac{\partial^{2}\eta}{\partial y^{2}}=0$
$\frac{\partial^{2}\eta}{\partial x\partial y}=\frac{\partial g}{\partial x}$
$\frac{\partial^{2}u}{\partial y^{2}}=Ug^{2}f^{\prime\prime\prime}$

$f^{\prime\prime}=\frac{\partial^{2}f}{\partial\eta ^{2}}$
$f^{\prime\prime\prime}=\frac{\partial^{3}f}{\parti al\eta^{3}}$

I'm leaving $\frac{\partial \eta}{\partial x}$ as it is because it should cancel out when the correct partials are plugged back into the actual aerodynamic problem.

That's what I've got on my own, $\frac{\partial^{2}u}{\partial y^{2}}$ is the one I'm pretty sure I'm lost on but I would be I'm probably off on others as well.

Thanks for the response thus far and any further help.